Problem: The Grunters play the Screamers 4 times.  The Grunters are the much better team, and are $75\%$ likely to win any given game.  What is the probability that the Grunters will win all 4 games? Express your answer as a common fraction.
Each of the 4 games is independent of the others, and in each game, the Grunters have probability $\frac34$ of winning.  Therefore, to get the probability that the Grunters will win all 4 games, we multiply the probabilities that the Grunters win each individual game.  This gives:  \begin{align*}
&P(\text{Grunters win all 4 games}) \\
&\quad= P(\text{Grunters win Game 1}) \times \cdots \times P(\text{Grunters win Game 4}) \\
&\quad= \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} \\
&\quad= \left(\frac{3}{4}\right)^{\!4} = \boxed{\frac{81}{256}}.
\end{align*}